Section Formula in 3D

What is the Section Formula in 3D?

The section formula in 3D geometry is a mathematical formula used to determine the coordinates of a point that divides a line segment joining two points in three-dimensional space in a given ratio. It is one of the most important concepts in coordinate geometry, vector algebra, and analytical geometry. The 3D section formula helps solve problems involving line segments, ratios, distances, midpoints, and spatial coordinates.

Section Formula Meaning in Simple Words

In simple words, the section formula helps find the exact coordinates of a point lying between or outside two points in three-dimensional space.

For example, if a point divides the line joining two points in the ratio $2:3$, the section formula can be used to determine its coordinates directly without plotting the points.

Definition of Section Formula in Three-Dimensional Geometry

The section formula in three-dimensional geometry is used to find the coordinates of a point that divides the line segment joining two given points in a specified ratio.

Suppose: $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ are two points in space.

If a point $P$ divides the line segment $AB$ in the ratio $m:n$, then the coordinates of $P$ can be calculated using the section formula.

Now:

If the point $R(x, y, z)$ divides the line joining the points $P\left(x_1, y_1, z_1\right)$ and $Q\left(x_2, y_2, z_2\right)$ internally in the ratio $m: n$, then the coordinate of $R(x, y, z)$ is given by :

$\mathrm{R}(\mathrm{x}, \mathrm{y} . \mathrm{z})=\left(\frac{m x_2+n x_1}{m+n}, \frac{m y_2+n y_1}{m+n}, \frac{m z_2+n z_1}{m+n}\right)$

If the point $R$ divides $P Q$ externally in the ratio $m$ : $n$, then its coordinates are obtained by replacing $n$ with $(-n)$, so that the coordinates of point R will be

$\mathrm{R}(\mathrm{x}, \mathrm{y}, \mathrm{z})=\left(\frac{m x_2-n x_1}{m-n}, \frac{m y_2-n y_1}{m-n}, \frac{m z_2-n z_1}{m-n}\right)$

NOTE:

1. The coordinates of the midpoint ( $\mathrm{m}: \mathrm{n}=1: 1$ ) of the line joining the points $P\left(x_1, y_1, z_1\right)$ and $Q\left(x_2, y_2\right.$, $z_2$ ) are as:

$\mathrm{R}(\mathrm{x}, \mathrm{y}, \mathrm{z})=\left(\frac{x_2+x_1}{2}, \frac{y_2+y_1}{2}, \frac{z_2+z_1}{2}\right)$

2. The coordinates of the point $R$ which divides $PQ$ in the ratio $k : 1$ are obtained by taking $\mathrm{k}=\mathrm{m} / \mathrm{n}$ which are as given below:

$\mathrm{R}(\mathrm{x}, \mathrm{y}, \mathrm{z})=\left(\frac{k x_2+x_1}{1+k}, \frac{k y_2+y_1}{1+k}, \frac{k z_2+z_1}{1+k}\right)$

3. The coordinates of the centroid of a triangle $R$ with vertices $A\left(x_1, y_1, z_1\right) B\left(x_2, y_2, z_2\right)$, and $C\left(x_3, y_3, z_3\right)$ are

$\mathrm{R}(\mathrm{x}, \mathrm{y}, \mathrm{z})=\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$

Best Books for Section Formula in 3D

A strong understanding of three-dimensional geometry and coordinate geometry is essential for mastering the section formula in 3D. The following books provide detailed explanations, derivations, and exam-oriented problems.

Shortcut Tips and Tricks for Section Formula in 3D

Understanding a few key shortcuts can help solve section formula questions quickly without lengthy calculations.

Important Formula Table

This formula table summarizes the most important formulas related to the section formula in three-dimensional geometry.

Solved Examples based on Section Formula in 3D

Example 1: Let the position vectors of two points $P$ and $Q$ be $3\hat{i}-\hat{j}+2\hat{k}$ and $\hat{i}+2\hat{j}-4\hat{k}$, respectively. Let $R$ and $S$ be two points such that the direction ratios of lines $PR$ and $QS$ are $(4,-1,2)$ and $(-2,1,-2)$, respectively. Let lines $PR$ and $QS$ intersect at $T$. If the vector $\overrightarrow{TA}$ is perpendicular to both $\overrightarrow{PR}$ and $\overrightarrow{QS}$ and the length of $\overrightarrow{TA}$ is $\sqrt{5}$ units, then find the modulus of the position vector of $A$.

Solution:

$P=(3,-1,2)$

$Q=(1,2,-4)$

$\overrightarrow{PR}\parallel(4\hat{i}-\hat{j}+2\hat{k})$

$\overrightarrow{QS}\parallel(-2\hat{i}+\hat{j}-2\hat{k})$

The direction ratios of the normal to the plane containing $P$, $T$, and $Q$ are proportional to

$\frac{l}{0}=\frac{m}{4}=\frac{n}{2}$

For point $T$,

$\frac{x-3}{4}=\frac{y+1}{-1}=\frac{z-2}{2}=\lambda$

and

$\frac{x-1}{-2}=\frac{y-2}{1}=\frac{z+4}{-2}=\mu$

Therefore,

$(4\lambda+3,,-\lambda-1,,2\lambda+2)\equiv(-2\mu+1,,\mu+2,,-2\mu-4)$

From the coordinates,

$4\lambda+3=-2\mu+1$

$\Rightarrow 2\lambda+\mu=-1$

and

$-\lambda-1=\mu+2$

$\Rightarrow \lambda+\mu=-3$

Solving,

$\lambda=2,\quad \mu=-5$

Hence,

$T=(11,-3,6)$

Now,

$\overrightarrow{OA}=(11\hat{i}-3\hat{j}+6\hat{k})\pm\left(\frac{2\hat{j}+\hat{k}}{\sqrt{5}}\right)\sqrt{5}$

$\overrightarrow{OA}=(11\hat{i}-3\hat{j}+6\hat{k})\pm(2\hat{j}+\hat{k})$

Thus,

$\overrightarrow{OA}=11\hat{i}-\hat{j}+7\hat{k}$

or

$\overrightarrow{OA}=11\hat{i}-5\hat{j}+5\hat{k}$

Therefore,

$|\overrightarrow{OA}|=\sqrt{11^2+(-1)^2+7^2}$

$=\sqrt{121+1+49}$

$=\sqrt{171}$

Hence, the answer is $\sqrt{171}$.

Example 2: If a unit vector $\vec{a}$ makes angles $\frac{\pi}{3}$ with $\hat{i}$, $\frac{\pi}{4}$ with $\hat{j}$ and $\theta \in (0,\pi)$ with $\hat{k}$, then a value of $\theta$ is:

[JEE Main 2019]

Solution:

For a unit vector, the direction cosines satisfy:

$l=\cos\alpha,\quad m=\cos\beta,\quad n=\cos\gamma$

and

$\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$

Substituting the given values:

$\cos^2\frac{\pi}{3}+\cos^2\frac{\pi}{4}+\cos^2\theta=1$

$\left(\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{2}}\right)^2+\cos^2\theta=1$

$\frac{1}{4}+\frac{1}{2}+\cos^2\theta=1$

$\cos^2\theta=\frac{1}{4}$

$\cos\theta=\pm\frac{1}{2}$

Therefore,

$\theta=\frac{\pi}{3}$

or

$\theta=\pi-\frac{\pi}{3}=\frac{2\pi}{3}$

Hence, the answer is $\frac{\pi}{3}$ or $\frac{2\pi}{3}$.

Example 3: Let $A(3,0,-1)$, $B(2,10,6)$ and $C(1,2,1)$ be the vertices of a triangle and $M$ be the midpoint of $AC$. If $G$ divides $BM$ in the ratio $2:1$, then $\cos(\angle GOA)$ (where $O$ is the origin) is equal to:

[JEE Main 2019]

Solution:

Since $M$ is the midpoint of $AC$ and $G$ divides $BM$ in the ratio $2:1$, point $G$ is the centroid of triangle $ABC$.

The coordinates of the centroid are:

$G=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3},\frac{z_1+z_2+z_3}{3}\right)$

Substituting the coordinates:

$G=\left(\frac{3+2+1}{3},\frac{0+10+2}{3},\frac{-1+6+1}{3}\right)$

$G=(2,4,2)$

Position vectors:

$\overrightarrow{OA}=3\hat{i}+0\hat{j}-\hat{k}$

$\overrightarrow{OG}=2\hat{i}+4\hat{j}+2\hat{k}$

Using the angle formula:

$\cos\theta=\frac{\overrightarrow{OG}\cdot\overrightarrow{OA}}{|\overrightarrow{OG}|,|\overrightarrow{OA}|}$

$\cos(\angle GOA)=\frac{(2)(3)+(4)(0)+(2)(-1)}{\sqrt{2^2+4^2+2^2}\sqrt{3^2+0^2+(-1)^2}}$

$=\frac{6-2}{\sqrt{24}\sqrt{10}}$

$=\frac{4}{2\sqrt{6}\sqrt{10}}$

$=\frac{1}{\sqrt{15}}$

Hence, the answer is $\frac{1}{\sqrt{15}}$

Example 4: An angle between the lines whose direction cosines are given by the equations $l+3m+5n=0$ and $5lm-2mn+6nl=0$ is:

[JEE Main 2018]

Solution:

The direction cosines satisfy:

$l^2+m^2+n^2=1$

Given:

$l+3m+5n=0$

$\Rightarrow l=-(3m+5n)$

Substituting into

$5lm-2mn+6nl=0$

we get:

$5(-(3m+5n))m-2mn+6n(-(3m+5n))=0$

$-15m^2-25mn-2mn-18mn-30n^2=0$

$15m^2+45mn+30n^2=0$

$m^2+3mn+2n^2=0$

$(m+n)(m+2n)=0$

Therefore,

$m=-n$

or

$m=-2n$

Taking $m=-n$,

$l=-(3(-n)+5n)=-2n$

Using

$l^2+m^2+n^2=1$

$(-2n)^2+(-n)^2+n^2=1$

$6n^2=1$

$n=\pm\frac{1}{\sqrt6}$

One set of direction cosines becomes

$\left(\frac{2}{\sqrt6},\frac{1}{\sqrt6},-\frac{1}{\sqrt6}\right)$

and the other becomes

$\left(\frac{1}{\sqrt6},-\frac{2}{\sqrt6},\frac{1}{\sqrt6}\right)$

Using the angle formula:

$\cos\theta=\frac{2}{6}-\frac{2}{6}-\frac{1}{6}$

$\cos\theta=-\frac{1}{6}$

Hence, the acute angle between the lines is

$\theta=\cos^{-1}\left(\frac{1}{6}\right)$

Hence, the answer is $\cos^{-1}\left(\frac{1}{6}\right)$.

Example 5: If the position vectors of the vertices $A$, $B$, and $C$ of a triangle $ABC$ are respectively $4\hat{i}+7\hat{j}+8\hat{k}$, $2\hat{i}+3\hat{j}+4\hat{k}$ and $2\hat{i}+5\hat{j}+7\hat{k}$, then find the position vector of the point where the bisector of $\angle A$ meets $BC$.

[JEE Main 2018]

Solution:

Let the angle bisector of $\angle A$ meet side $BC$ at point $P(x,y,z)$.

Using the Angle Bisector Theorem,

$\frac{BP}{PC}=\frac{AB}{AC}$

Calculating lengths:

$AB=\sqrt{(4-2)^2+(7-3)^2+(8-4)^2}$

$=\sqrt{4+16+16}$

$=6$

$AC=\sqrt{(4-2)^2+(7-5)^2+(8-7)^2}$

$=\sqrt{4+4+1}$

$=3$

Therefore,

$AB:AC=6:3=2:1$

Hence,

$BP:PC=2:1$

Point $B=(2,3,4)$ and point $C=(2,5,7)$.

Using the section formula,

$P=\left(\frac{2(2)+1(2)}{3},\frac{2(5)+1(3)}{3},\frac{2(7)+1(4)}{3}\right)$

$=\left(\frac{6}{3},\frac{13}{3},\frac{18}{3}\right)$

$=\left(2,\frac{13}{3},6\right)$

Therefore, the position vector of $P$ is

$\frac{1}{3}(6\hat{i}+13\hat{j}+18\hat{k})$

Hence, the answer is $\frac{1}{3}(6\hat{i}+13\hat{j}+18\hat{k})$.

Related Topics to Section Formula in 3D

Understanding the section formula in 3D becomes much easier when studied alongside other coordinate geometry concepts such as distance formula, midpoint formula, direction ratios, vectors, and three-dimensional geometry. Exploring these related topics helps build a strong foundation for solving advanced geometry and analytical mathematics problems in board exams and competitive examinations.